If $(X_n)_{n\geq 1}$ is a submartingale and $S\leq T$ are bounded stopping times, then $\mathbb{E}(X_T)\geq\mathbb{E}(X_S)$.
I am quite new to conditional expectation, martingales etc. In our lecture the professor mentioned that the above can be proved by using the Doob Decomposition Theorem, however I don't quite see it. Could someone please explain this to me. Would be greatly appreciated, thank you in advance.
My attempt:
$$\begin{align*} \mathbb{E}(X_T)=\mathbb{E}(X_1+M_T+A_T) &=\mathbb{E}(X_1)+\mathbb{E}(M_T|\mathcal{F}_S)+\mathbb{E}(A_T)\\&=\mathbb{E}(X_1)+\mathbb{E}(\mathbb{E}(M_T|\mathcal{F}_S))+\mathbb{E}(A_T)\end{align*}$$ Then since $\mathbb{E}(M_T)=\mathbb{E}(M_0)$ for any martingale $M_n$ and any bounded stopping time T, we have $\mathbb{E}(\mathbb{E}(M_T|\mathcal{F}_S))=\mathbb{E}(M_0)$, so $$\begin{align*}&=\mathbb{E}(X_1)+\mathbb{E}(M_0)+\mathbb{E}(A_T)\end{align*}$$ Now since $M_S$ is also a martingale and S is also a bounded stopping time, we again have $\mathbb{E}(M_0)=\mathbb{E}(M_S)$, $$\begin{align*}&=\mathbb{E}(X_1)+\mathbb{E}(M_S)+\mathbb{E}(A_T)\end{align*}$$ Finally, since according to Doob's Decomposition theorem $(A_n)_{n\geq 0}$ is an increasing predictable process, $$\begin{align*}\mathbb{E}(X_1)+\mathbb{E}(M_S)+\mathbb{E}(A_T)&\geq\mathbb{E}(X_1)+\mathbb{E}(M_S)+\mathbb{E}(A_S)\\ &=\mathbb{E}(X_S)\end{align*}$$